Question

The velocity of a 3.00 kg parti- cle is given by :v = (8.00tiˆ + 3.00t2jˆ) m/s, with time t in seconds. At the instant the net force on the parti- cle has a magnitude of 35.0 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?

Answer 1

Answer:

Part a)

Direction of net force is

$\theta = 46.7 degree$

Part b)

Direction of the velocity is given as

$\phi = 27.9 degree$

Explanation:

As we know that the velocity of the particle is given as

$v = 8.00 t \hat i + 3.00 t^2\hat j$

now the acceleration is given as

$a = \frac{dv}{dt}$

$a = 8.00 \hat i + 6.00 t\hat j$

now magnitude of net acceleration is given as

$a = \sqrt{64 + 36t^2}$

$F = 3a$

$35 = 3\sqrt{64 + 36t^2}$

$t = 1.41 s$

Part a)

Now direction of net force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{6t}{8}$

$tan\theta = \frac{6(1.41)}{8}$

$\theta = 46.7 degree$

Part b)

Direction of the velocity is given as

$tan\phi = \frac{v_y}{v_x}$

$tan\phi = \frac{3.00 t^2}{8.00 t}$

$tan\phi = \frac{3.00(1.41)}{8.00}$

$\phi = 27.9 degree$